Question

Consider the vaporization of a typical molecular liquid at 200 atm? The specific volumes of this molecule in this liquid and gas phases are 1.000 and 3.0000 cm³/g at 100 degrees C, respectively. The liquid boils at 100 degrees C at 1 atm. By how much does the boiling point of this liquid change in response to the application of pressure?

Answer 1

Apply Clausius- Clayperon eqquation

T_B = [1/ To - (R ln P/P_o/ H_vap)]^-1 ------------------------- (1)

T_{B =} Boiling point at pressure of ineterest

To = Normal boiling temperature = 100^oC

Po = Pressure at normal boiling point = 1 atm = 101325 Pa

P = increased pressure = 200 atm

R = 8.314 J/molK

_vap = enthalpy of vaporization = p = work done against ambient pressure

Consider weight of liquid molecule = 100 kg

= Change in specific volume (m³/kg) x weight (kg)

_vap= 101325 x (3 - 1)*10^-3 * 100 = 20265 J (1 cm³/ g = 10^-6/ 10^-3m³/ kg = 10^-3m³/ kg)

Put all values in (1)

T_B = [0.01 - (8.314 ln200/ 20265)]^-1 = [0.01 - 0.00217]^-1 = 0.00783^-1

T_B = 127.71^oC